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Bernoulli's Problem

  1. Jul 25, 2009 #1

    jdg

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    Next question: A garden hose with internal diameter of 13.5 mm lies flat on a sidewalk while water is flowing in it at a speed of 6 m/s. A person happens to step on it at the very edge of the opening of the hose and decreases its internal diameter by a factor of 9

    So D (1) = 0.0135m
    r (1) = 0.00675m
    D (2) = 0.0135/9 = 0.0015m
    r (2) = 0.00075m
    A (1) = pi*r^2 = (3.14...)(0.00675^2) = 1.4134...e-4
    A (2) = pi*r^2 = (3.14...)(0.00075^2) = 1.767...e-6

    1. What was the water flow rate in the hose prior to the person stepping on it?
    - I got this part: J(1) = A(1)V(1) = 8.59 m3/s

    2. What is the flow rate of water after the person steps on it?

    3. What is the speed of the water just as it exits the hose after the person steps on it?
     
  2. jcsd
  3. Jul 25, 2009 #2
    This actually is not a Bernoulli's equation problem.

    Check your textbook (especially the section that I assume includes this problem) for something called a 'Continuity Equation'. Continuity of mass for an incompressible fluid says that the flow rate is a constant. That fact should be enough to get you through Parts 2 and 3.
     
  4. Jul 25, 2009 #3

    jdg

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    Ok, for Q2, part 2 I did J = A(1)V(1) = A(2)V(2):

    So V2 = V1*(A1/A2) = 486 m/s

    Is this right?

    And for part 3 I did

    J = (A2)(V2) = 8.59e-4 m3/s
     
  5. Jul 25, 2009 #4
    Flow rate = Av = constant.

    Which means flow rate before the hose is stepped on = flow rate after the hose is stepped on.

    Which means, if you are doing the problem correctly, what you are calling "J" should be the same bfore the hose is stepped on and after the hose is stepped on.
     
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